Animated Solution for Chemistry - Chemical Equilibrium: Which of the following lines correctly show the temperature dependence of equilibrium constant, $K$, for an exothermic reaction?

Imagine you are a chemist trying to maximize the yield of a crucial industrial reaction. You know that changing the temperature shifts the equilibrium, but how exactly does it happen? Does the equilibrium constant $K$ go up or down? This is where the beautiful intersection of thermodynamics and chemical equilibrium comes into play.

In this problem, we are presented with a graph plotting the natural logarithm of the equilibrium constant, $\ln K$, against the inverse of temperature, $\frac{1}{T}$. We are asked to identify which of the four lines ($A,B,C,$ or $D$) correctly represents an exothermic reaction. Let's break down the physics and math behind this.

To understand the graph, we need a mathematical bridge that connects our y-axis ($\ln K$) to our x-axis ($\frac{1}{T}$). That bridge is the famous van 't Hoff equation, derived from the fundamental principles of Gibbs free energy:

Here, $\Delta H^{\circ }$ is the standard enthalpy change, $\Delta S^{\circ }$ is the standard entropy change, and $R$ is the universal gas constant. This equation is incredibly powerful because it reveals exactly how temperature influences the equilibrium state of a reaction.

Decoding the Straight Line

Let's look closely at the structure of the van 't Hoff equation. It perfectly mirrors the standard equation of a straight line:

By mapping our variables, we get: - y-axis: $y=\ln K$ - x-axis: $x=\frac{1}{T}$ - Slope ($m$): $m=-\frac{\Delta H^{\circ }}{R}$ - y-intercept ($c$): $c=\frac{\Delta S^{\circ }}{R}$

This mapping is the key to unlocking the graph. The slope of any line on this plot is directly proportional to the negative of the enthalpy change, and the y-intercept is directly proportional to the entropy change.

The Heat is On

The question specifically asks us to identify the lines corresponding to an exothermic reaction. What defines an exothermic reaction? It is a process that releases heat into its surroundings, which means its standard enthalpy change is strictly negative:

Now, let's substitute this negative value into our slope formula:

Because a negative multiplied by a negative yields a positive, the slope $m$ must be positive ($m>0$).

Looking back at our graph, we need to find the lines that are going upwards from left to right. These are lines A and B. Both of these lines have a positive slope, perfectly satisfying the condition for an exothermic reaction.

The Hidden Intercept

You might wonder, why are there two lines (A and B) for an exothermic reaction? The answer lies in the y-intercept, which represents the entropy change, $\frac{\Delta S^{\circ }}{R}$.

- Line A has a positive y-intercept, meaning $\Delta S^{\circ }>0$. This represents an exothermic reaction where the system's disorder increases (e.g., a solid turning into a gas). - Line B has a negative y-intercept, meaning $\Delta S^{\circ }<0$. This represents an exothermic reaction where the system becomes more ordered (e.g., two gas molecules combining into one).

Both scenarios are physically possible, which is why both lines A and B are correct representations.

Bringing It All Together

By leveraging the van 't Hoff equation, we translated a thermodynamic concept into a simple geometric property: the slope of a line. We found that exothermic reactions always produce a positive slope on a $\ln K$ vs $\frac{1}{T}$ plot.

Conversely, lines C and D have negative slopes, which means they represent endothermic reactions ($\Delta H^{\circ }>0$). Always remember this visual shortcut: if the line goes up, heat is given off; if the line goes down, heat is taken in. This intuitive understanding will serve you well in mastering chemical equilibrium!